Answer :
The molarity is given as 16 M, the volume of 16 M [tex]NH3[/tex] solution required to prepare 2.00 L of a 2.00 M solution is 0.25 liters or 250 ml.
To prepare a 2.00 M solution of [tex]NH3[/tex] (ammonia) with a volume of 2.00 L, we can use the formula for calculating the amount of solute needed:
moles of solute = Molarity × Volume (in liters)
Molarity of the desired solution (M1) = 2.00 M
Volume of the desired solution (V1) = 2.00 L
Molarity of the stock [tex]NH3[/tex] solution (M2) = 16 M
Let's calculate the moles of [tex]NH3[/tex] needed:
moles of [tex]NH3[/tex] = M1 × V1
moles of [tex]NH3[/tex] = 2.00 M × 2.00 L
moles of [tex]NH3[/tex] = 4.00 moles
Now, we need to calculate the volume of the 16 M [tex]NH3[/tex] solution containing 4.00 moles of [tex]NH3[/tex]:
Volume of 16 M [tex]NH3[/tex]solution (V2) = moles of [tex]NH3[/tex] ÷ M2
Volume of 16 M [tex]NH3[/tex] solution (V2) = 4.00 moles ÷ 16 M
Volume of 16 M [tex]NH3[/tex] solution (V2) = 0.25 L
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